Question

a 50.0 g sample of water absorbs 1050 j of heat. if the specific heat capacity of water is 4.18 j/g°c, what is the change in temperature of the water?
5.02 °c
52.5 °c
21.0 °c
25.1 °c

during an experiment, a student observes that 836 j of heat are released when a 40.0 g sample of an unknown substance cools by 10.0°c. what is the specific heat capacity of this substance?
0.21 j/g°c
8.36 j/g°c
334.4 j/g°c
2.09 j/g°c

Explanation:

Step1: Recall heat - capacity formula

The formula for heat transfer is $Q = mc\Delta T$, where $Q$ is the heat absorbed or released, $m$ is the mass of the substance, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. We can re - arrange it to solve for $\Delta T$ as $\Delta T=\frac{Q}{mc}$ and for $c$ as $c = \frac{Q}{m\Delta T}$.

Step2: Solve for the change in temperature of water

Given $Q = 1050\ J$, $m = 50.0\ g$, and $c = 4.18\ J/g^{\circ}C$. Substitute these values into the formula $\Delta T=\frac{Q}{mc}$.
$\Delta T=\frac{1050\ J}{50.0\ g\times4.18\ J/g^{\circ}C}=\frac{1050}{209}\ ^{\circ}C\approx5.02\ ^{\circ}C$

Step3: Solve for the specific heat capacity of the unknown substance

Given $Q=- 836\ J$ (negative because heat is released), $m = 40.0\ g$, and $\Delta T=-10.0^{\circ}C$ (negative because the substance cools). Substitute into the formula $c=\frac{Q}{m\Delta T}$.
$c=\frac{-836\ J}{40.0\ g\times(- 10.0^{\circ}C)}=\frac{-836}{-400}\ J/g^{\circ}C = 2.09\ J/g^{\circ}C$

Answer:

1. A. $5.02\ ^{\circ}C$
2. D. $2.09\ J/g^{\circ}C$